Let n ≥ 5 be an integer. If 9ⁿ - 8n - 1 = 64α and 6ⁿ - 5n - 1 = 25β, then α - β is equal to: (A) 1 +

Q: Wait, why do we write $9$ as $(1+8)$ instead of $(3^2)$? Wouldn't that be easier?

That is a fantastic question! While $9$ is indeed $3^2$, the Binomial Theorem is designed to expand expressions of the form $(1+x)^n$. By writing $9$ as $(1+8)$, we align the expression perfectly with the structure of the Binomial Theorem, allowing us to isolate the terms we need. If we used $3^2$, we would be stuck with powers of $3$, which doesn't help us isolate the coefficients in the way the problem requires.

Q: Why does the ${}^nC_2$ term cancel out in the final subtraction?

It is not that it cancels out to zero, but rather that the ${}^nC_2$ term in the expression for $\alpha$ and the ${}^nC_2$ term in the expression for $\beta$ are identical. When we calculate $\alpha - \beta$, we are subtracting the entire series. Since both series start with the constant term ${}^nC_2$, they subtract to zero, leaving us with the higher-order terms starting from ${}^nC_3$.

Q: Is there a general rule for these types of problems?

Yes! Whenever you see an expression like $(1+x)^n - nx - 1$, you should immediately think of the Binomial Theorem. This specific form is always divisible by $x^2$. Recognizing this pattern is a hallmark of a JEE Advanced aspirant. It turns a seemingly complex algebraic problem into a straightforward series manipulation.

Question

Let $n \ge 5$ be an integer. If $9^n - 8n - 1 = 64\alpha$ and $6^n - 5n - 1 = 25\beta$, then $\alpha - \beta$ is equal to:

Options:

(A) $1 + {}^{n}C_{2}(8-5) + {}^{n}C_{3}(8^{2}-5^{2}) + \dots + {}^{n}C_{n}(8^{n-1}-5^{n-1})$

(B) $1 + {}^{n}C_{3}(8-5) + {}^{n}C_{4}(8^{2}-5^{2}) + \dots + {}^{n}C_{n}(8^{n-2}-5^{n-2})$

(C) ${}^{n}C_{3}(8-5) + {}^{n}C_{4}(8^{2}-5^{2}) + \dots + {}^{n}C_{n}(8^{n-2}-5^{n-2})$

(D) ${}^{n}C_{4}(8-5) + {}^{n}C_{5}(8^{2}-5^{2}) + \dots + {}^{n}C_{n}(8^{n-3}-5^{n-3})$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path to JEE mastery. Today, we are not just solving an equation; we are peeling back the layers of a mathematical onion to reveal the elegant structure hidden beneath.

We are given two expressions: 
$$9^n - 8n - 1 = 64\alpha$$ 
and 
$$6^n - 5n - 1 = 25\beta$$

The terms $-8n - 1$ and $-5n - 1$ are clear indicators to apply the **Binomial Theorem**. Recall the expansion: 
$$(1+x)^n = 1 + nx + {}^nC_2 x^2 + {}^nC_3 x^3 + \dots + {}^nC_n x^n$$

By rewriting $9$ as $(1+8)$ and $6$ as $(1+5)$, we transform these expressions into a form where the Binomial Theorem can do the heavy lifting for us.

### The Art of Isolation

Let us focus on the first expression. We write $9^n$ as $(1+8)^n$. Expanding this using our theorem, we get:
$$(1+8)^n = 1 + n(8) + {}^nC_2(8^2) + {}^nC_3(8^3) + \dots + {}^nC_n(8^n)$$

Subtracting $1$ and $8n$ from both sides, we are left with the terms starting from ${}^nC_2(8^2)$:
$$64\alpha = {}^nC_2(8^2) + {}^nC_3(8^3) + \dots + {}^nC_n(8^n)$$

To isolate $\alpha$, we divide by $64$ (which is $8^2$). The powers of $8$ reduce beautifully:
$$\alpha = {}^nC_2 + {}^nC_3(8) + {}^nC_4(8^2) + \dots + {}^nC_n(8^{n-2})$$

### The Grand Cancellation

Now, we repeat this logic for $\beta$. Writing $6^n$ as $(1+5)^n$ and following the exact same steps, we find:
$$\beta = {}^nC_2 + {}^nC_3(5) + {}^nC_4(5^2) + \dots + {}^nC_n(5^{n-2})$$

Now, we arrive at the climax of our journey: calculating $\alpha - \beta$. When we subtract these two series term-by-term, the first term, ${}^nC_2$, appears in both and cancels out completely.

We are left with the final expression:
$$\alpha - \beta = {}^nC_3(8-5) + {}^nC_4(8^2-5^2) + \dots + {}^nC_n(8^{n-2}-5^{n-2})$$

This is the beauty of mathematics—the way complex terms vanish, leaving behind a clean, elegant result. You have not just solved a problem; you have mastered the art of pattern recognition.

Animated Solution for Mathematics - Binomial Theorem: Let $n \geq 5$ be an integer. If $9^{n} - 8 n - 1 = 64 α$ and $6^{n} - 5 n - 1 = 25 β$ , then $α - β$ is equal to:

Select Answer:

Visualized Solution

Given Equations

The Binomial Theorem

Expanding $9^{n}$

Isolating $64 α$

Finding the Series for $α$

Expanding $6^{n}$

Finding the Series for $β$

Calculating $α - β$

Final Result

The Sigma Insight: Binomial Expansion for Positive Integral Index

Analyzing the Setup

The Art of Isolation

The Grand Cancellation

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Animated Solution for Mathematics - Binomial Theorem: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let n≥5 be an integer. If 9n−8n−1=64α and 6n−5n−1=25β, then α−β is equal to:

Select Answer:

Visualized Solution

Given Equations

The Binomial Theorem

Expanding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 9n

Isolating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 64α

Finding the Series for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α

Expanding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 6n

Finding the Series for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } β

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α−β